# Orthogonal Matrices - Examples with Solutions

  

## Definition of Orthogonal Matrices

An $n \times n$ matrix whose columns form an orthonormal set is called an orthogonal matrix.
As a reminder, a set of vectors is orthonormal if each vector is a unit vector ( length or norm of the vector is equal to $1$) and each vector in the set is orthogonal to all other vectors in the set.
These are examples of orthogonal matrices.
a) $\begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & 1 \\ 0 & -1 & 0\\ \end{bmatrix}$        b) $\begin{bmatrix} \dfrac{1}{\sqrt {10}} & 0 & - \dfrac{3}{\sqrt {10}}\\ \dfrac{3}{\sqrt {10}} & 0 & \dfrac{1}{\sqrt {10}} \\ 0 & -1 & 0\\ \end{bmatrix}$        c) $\begin{bmatrix} \dfrac{1}{\sqrt{6}}&0&\dfrac{5}{\sqrt{30}}\\ \:\dfrac{1}{\sqrt{6}}&-\dfrac{2}{\sqrt{5}}&-\dfrac{1}{\sqrt{30}}\\ \:\dfrac{2}{\sqrt{6}}&\dfrac{1}{\sqrt{5}}&-\dfrac{2}{\sqrt{30}} \end{bmatrix}$        d) $\begin{bmatrix} \cos x& -\sin x \\ \sin x&\cos x \end{bmatrix}$

## Properties of Orthogonal Matrices

A list of the most important properties of orthogonal matrices is given below. If $Q$ is an orthogonal matrix, then

1.   $Q^{-1} = Q^T$ ; this the most important property of orthogonal matrices as the inverse is simply the transpose.
2.   the rows of $Q$ form an orthonormal set.
3.   $Q^{-1}$ is an orthogonal matrix
4.   det $( Q ) = \pm 1$
5.   if $\lambda$ is an eigrnvalue of Q, then $|\lambda| = 1$
6.   if $Q_1$ and $Q_2$ are $n \times n$ orthogonal matrices, then $Q_1 Q_2$ is also an orthogonal matrix.

## Examples with Solutions

Example 1
The matrices $Q_1 = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$ and $Q_2 = \begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & -1 \end{bmatrix}$ are orthogonal. Verify that the product $Q_1 Q_2$ is also orthogonal (Property 6 above)

Solution
We first calculate the product $Q_1 Q_2$
$Q_1 Q_2 = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & -1 \end{bmatrix} = \begin{bmatrix} 0&0&-1\\ 0&-1&0\\ 1&0&0 \end{bmatrix}$
Let $\textbf v_1 , \textbf v_2 , \textbf v_3$ be the columns of the matrix $Q_1 Q_2$ found above.
$\textbf {v}_1 = \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}$ , $\textbf {v}_2 = \begin{bmatrix} 0\\ -1\\ 0 \end{bmatrix}$ , $\textbf {v}_3 = \begin{bmatrix} -1\\ 0\\ 0 \end{bmatrix}$
Let us calculate the length or norm of each column
$|| \textbf {v}_1 || = \sqrt {0^2+0^2+1^2} = 1$
$|| \textbf {v}_2 || = \sqrt {0^2+(-1)^2+0^2} = 1$
$|| \textbf {v}_3 || = \sqrt {(-1)^2+0^2+0^2} = 1$
All three vectors are unit vectors.
Calculate the inner product of all pairs of vectors that can be made from the vectors $\textbf v_1 , \textbf v_2 , \textbf v_3$
$\textbf {v}_1 \cdot \textbf {v}_2 = 0 \cdot 0 + 0 \cdot (-1) + 1 \cdot 0 = 0$
$\textbf {v}_1 \cdot \textbf {v}_3 = 0 \cdot (-1) + 0 \cdot 0 + 1 \cdot 0 = 0$
$\textbf {v}_2 \cdot \textbf {v}_3 = 0 \cdot (-1) + (-1) \cdot 0 + 0 \cdot 0 = 0$
The three vectors form an orthogonal set.
The three columns of the matrix $Q_1 Q_2$ are orthogonal and have norm or length equal to 1 and are therefore orthonormal.

Example 2
Use a calculator to find the inverse of the orthogonal matrix matrix $Q = \begin{bmatrix} 0 & 0 & 1 \\ -1 & 0 & 0 \\ 0 & -1 & 0 \end{bmatrix}$ and verify Property 1 above.

Solution
Use any matrix calculator to find
$Q^{-1} = \begin{bmatrix} 0 & -1 & 0 \\ 0 & 0 & -1 \\ 1 & 0 & 0 \end{bmatrix}$
Find the transpose of matrix $Q$
$Q^T = \begin{bmatrix} 0 & -1 & 0 \\ 0 & 0 & -1 \\ 1 & 0 & 0 \end{bmatrix}$
Hence $Q^{-1} = Q^T$ , property 1 above.

Example 3
Find the real constants $a$ and $b$ in the matrix $Q = \begin{bmatrix} -1 & 0 & 0 \\ 0 & \dfrac{1}{\sqrt 2} & a \\ 0 & \dfrac{1}{\sqrt 2} & b \end{bmatrix}$ such that $Q$ is orthogonal.

Solution
Let $\textbf v_1 , \textbf v_2 , \textbf v_3$ be the columns of the matrix $Q$ given above such that
$\textbf {v}_1 = \begin{bmatrix} -1\\ 0\\ 0 \end{bmatrix}$ , $\textbf {v}_2 = \begin{bmatrix} 0\\ \dfrac{1}{\sqrt 2}\\ \dfrac{1}{\sqrt 2} \end{bmatrix}$ , $\textbf {v}_3 = \begin{bmatrix} 0\\ a\\ b \end{bmatrix}$
Two sets of conditions for matrix $Q$ to be orthogonal:
1) The norm of each column $\textbf v_1 , \textbf v_2 , \textbf v_3$ must equal to 1
$|| \textbf v_1 || = \sqrt {(-1)^2+0^2+0^2} = 1$
$|| \textbf v_2 || = \sqrt {0^2+ \left(\dfrac{1}{\sqrt 2} \right)^2+\left(\dfrac{1}{\sqrt 2} \right)^2 } = 1$
$|| \textbf v_3 || = \sqrt {0^2+a^2+b^2} = \sqrt {a^2+b^2} = 1$         (I)
2) The inner product of any two vectors must be equal to zero (orthogonal vetcors)
$\textbf {v}_1 \cdot \textbf {v}_2 = (-1) \cdot 0 + 0 \cdot \left(\dfrac{1}{\sqrt 2} \right) + 0 \cdot \left(\dfrac{1}{\sqrt 2} \right) = 0$
$\textbf {v}_1 \cdot \textbf {v}_3 = (-1) \cdot 0 + 0 \cdot a + 0 \cdot b = 0$
$\textbf {v}_2 \cdot \textbf {v}_3 = 0 \cdot 0 + \dfrac{1}{\sqrt 2} \cdot a + \dfrac{1}{\sqrt 2} \cdot b = \dfrac{1}{\sqrt 2} \cdot a + \dfrac{1}{\sqrt 2} \cdot b = 0$         (II)
For all conditions to be satisfied, we need to solve equations (I) and (II) above
Equation (II) above gives $a = - b$
Substitute $a$ by $- b$ in equation (I) to obtain
$\sqrt {2 b^2 } = 1$
Solve to obtain
$b = \pm \dfrac{1}{\sqrt 2}$
Two solutions to the above question
$a = - \dfrac{1}{\sqrt 2}$ and $b = \dfrac{1}{\sqrt 2}$
$a = \dfrac{1}{\sqrt 2}$ and $b = - \dfrac{1}{\sqrt 2}$

## Questions (with solutions given below)

• Part 1
Matrix $Q = \begin{bmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ is orthogonal. Verify the first 5 properties, listed above, for matrix $Q$.
• Part 2
1) Which of the following matrices are orthogonal?
a) $A = \begin{bmatrix} \dfrac{1}{\sqrt 2} & \dfrac{1}{\sqrt 2} \\ - \dfrac{1}{\sqrt 2} & \dfrac{1}{\sqrt 2} \end{bmatrix}$ , b) $B = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}$ , c) $C = \begin{bmatrix} \dfrac{1}{\sqrt {10}} & - \dfrac{3}{\sqrt {10}} & 0 \\ 0 & 0 & 1 \\ \dfrac{3}{\sqrt {10}} & \dfrac{1}{\sqrt {10}} & 0 \end{bmatrix}$
• Part 3
1) Find all matrices of the form $A = \begin{bmatrix} p & q\\ \dfrac{1}{\sqrt {3}} & r \end{bmatrix}$ that are orthogonal.
• Part 4
Find the inverse of matrix $A = \begin{bmatrix} 0 & \dfrac{5}{3 \sqrt 5} & \dfrac{2}{3} \\ \dfrac{1}{\sqrt 5} & - \dfrac{4}{3 \sqrt 5} & \dfrac{2}{3} \\ \dfrac{2}{\sqrt 5} & \dfrac{2}{3 \sqrt 5} & - \dfrac{1}{3} \end{bmatrix}$